Parseval's identity

[[concept]]

Topics

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Theorem (Parseval's Identity)

Let $H$ be a Hilbert space and let $\{e_n\}$ be a countable orthonormal basis of $H$. Then for all $u\in H$, ${\sum }_n|⟨u,e_n⟩{|}^2=||u|{|}^2$ (in Bessel’s inequality, we only had $\le$)

Proof

We know that $u={\sum }_n⟨u,e_n⟩e_n$ from all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements. If the sum over $n$ is a finite sum, then we just expand the inner product $||u|{|}^2=⟨u,u⟩$. Otherwise, by continuity of inner product, we have

\lvert \lvert u \rvert \rvert ^2 &= \lim_{ m \to \infty } \left\langle \sum_{n=1}^m \langle u, e_{n} \rangle e_{n}, \sum_{\ell=1}^m \langle u, e_{\ell} \rangle e_{\ell} \right\rangle \\ &= \lim_{ m \to \infty } \sum_{n, \ell = 1}^m\langle u, e_{n} \rangle \overline{\langle u, e_{\ell} \rangle } \langle e_{n, e_{\ell}} \rangle \\ (e_{n} \perp e_{\ell}, n \neq \ell) \implies&= \lim_{ m \to \infty } \sum_{n=1} \langle u, e_{n} \rangle \overline{\langle u, e_{n} \rangle } \\ &= \lim_{ m \to \infty } \sum_{n=1}^m \lvert \langle u, e_{n} \rangle \rvert ^2 \end{align}$$ $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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